# HW14Sol - τ 1 =(2.00m F 2 –(1.00m 196N –(1.70m 490.N =...

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Jim Guinn’s PHYS2211 Assignment #14 Solutions Due Wednesday, Mar. 31, 2010 1. A 2.00m plank with a mass of 20.0kg is supported at both ends. A person with a mass of 50.0kg stands at 1.70m from the left end. What are the forces which the supports exert on the plank? We have that the weight of the plank is given by W pl = m pl g = (20.0kg) (9.80m/s 2 ) = 196N , and the weight of the person is given by W pe = m pe g = (50.0kg) (9.80m/s 2 ) = 490.N . Setting the origin at the left support, and with counterclockwise as positive, we have the net torque acting on the plank is given by
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Unformatted text preview: τ 1 = (2.00m) F 2 – (1.00m) 196N – (1.70m) 490.N = I α = 0 , so (2.00m) F 2 = 196N m + 833N m = 1029N m , so F 2 = (1029N m) / (2.00m) = F 2 = 515N . Moving the origin to the right support we have the net torque is given by τ 2 = -(2.00m) F 1 + (1.00m) 196N + (0.30m) 490.N = I α = 0 , so (2.00m) F 1 = 196N m + 147N m = 343N m , so F 1 = (343N m) / (2.00m) = F 1 = 172N . Notice that F 1 + F 2 = W pl + W pe (just off by one at the last sig. fig.). F 1 F 2 W pl W pe 1.70m...
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## This note was uploaded on 09/25/2011 for the course PHYSICS 22 taught by Professor Lomant during the Spring '11 term at Georgia Perimeter.

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